Question

ПОМОГИТЕ ПОЖАЛУЙСТА АЛГЕБРА ДАЮ 100 БАЛОВ

Answer 1

1)

$f(x) = \frac{ {x}^{2} - 2 }{x} \\ f'(x) = \frac{( {x}^{2} - 2)'x - x '( {x}^{2} - 2) }{ {x}^{2} } = \\ \frac{2x \times x - ( {x}^{2} -2 ) }{ {x}^{2} } = \\ \frac{2 {x}^{2} - {x}^{2} + 2}{ {x}^{2} } = \frac{ {x}^{2} + 2 }{ {x}^{2} }$

2)

$f(x) = x \cos(x) \\ f'(x) = x' \cos(x) + (\cos(x) )' x= \\ \cos(x) -x \sin(x)$

3)

$f(x) = \frac{ \sqrt{x} + 2x }{x} \\ f'(x) = \frac{( \sqrt{x} + 2x)'x - x '( \sqrt{x} + 2x) }{ {x}^{2} } = \\ \frac{ (\frac{1}{2 \sqrt{x} } + 2)x - ( \sqrt{x} + 2x) }{ {x}^{2} } = \\ \frac{ \frac{x}{2 \sqrt{x} } + 2x - \sqrt{x} - 2x }{ {x}^{2} } = \frac{ \frac{x - \sqrt{x} \times 2 \sqrt{x} }{2 \sqrt{x} } }{ {x}^{2} } = \\ \frac{x - 2x}{2 \sqrt{x} \times {x}^{2} } = \frac{ - x}{2 \sqrt{x} \times {x}^{2} } = - \frac{1}{2x \sqrt{x} }$

4)

$f(x) = {x}^{2} \sin(x ) \\ f'(x) = ( {x}^{2} )' \sin(x) + {x}^{2} ( \sin(x) )' = \\ 2x \sin(x) + {x}^{2} \cos(x)$